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Equilibrium Question

  1. Nov 11, 2003 #1
    Two spheres of radius R rest on a horizontal table with their centers a distance 4R apart. One sphere has twice the weight of the other sphere. Where is the center of gravity of this system?

    Please help me solve this problem.
     
  2. jcsd
  3. Nov 11, 2003 #2

    HallsofIvy

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    Well, it's pretty hard to help you if you won't tell us what you do understand about this problem, what you have tried and where you run into a problem.

    To start with basics, what do you understand the definition of "center of gravity" to be?
     
  4. Nov 11, 2003 #3
    I think by center of gravity he meant the RADIUS in Fg.

    Fg = (G*m1*m2)/R2

    R = sqrt((G*m1*m2)/Fg)
     
  5. Nov 12, 2003 #4

    jamesrc

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    Actually, I think you want to approach this like a moment balance problem, where a force of 3Mg (the sum of the weights) is concentrated at the unknown x_cm (center of mass/gravity).
     
  6. Nov 12, 2003 #5

    HallsofIvy

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    Prudens Optimus, "You cannot be serious!" Actually, I thought for a very brief moment that he was talking about the gravitational attraction of the two spheres but that doesn't really make sense.

    jamesrc: I actually do know what "center of gravity" means! I asked timtng because I wanted to see what HIS understanding of the problem was.


    timtng: In other words, shorn of all the unecessary "sphere" stuff, you have a line supporting two weights of mass M and 2M with distance 4R between them and want to find a "fulcrum point" where they will balance. Let x be the distance from mass M to the fulcrum. Mass M has weight Mg and so torque Mgx about the fulcrum. Mass 2M has weight 2Mg. It's distance from the fulcrum is 4R- x so its torque around the fulcrum is 2Mg(4R-x). In order to balance those must be equal: Mgx= 2Mg(4R-x). The first thing you can do is divide the equation by Mg to get x= 2(4R- x)= 8R- 2x. 3x= 8R so x= 8R/3.
     
  7. Nov 12, 2003 #6

    jamesrc

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    HallsofIvy: I had no doubt of your understanding of the topic; what you were trying to do was quite clear. I just didn't want timtng to remain confused once gravitational attraction was brought into the discussion.
     
  8. Nov 12, 2003 #7

    HallsofIvy

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    Ah, I see your point. Thank you.
     
  9. Nov 13, 2003 #8
    why you people questioning timtng question....

    he may not be confused...what he means is really center of gravity...
    center of gravity not necessarily taken just from one object..

    in this case the

    assume take point O at the center gravity of smaller sphere.
    let w denotes the weight of smaller sphere.

    so the Center gravity of the total body =2w*4r / 3w = (8/3)r from point O

    it means the center of gravity lies on (5/3)r from lighter sphere or (1/3)r from weighter sphere and this center of gravity should lay in the line which connect the center gravity of lighter sphere and the center gravity from the weight one.

    I wish this answer and explanation,will benefits you
     
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